Answer:
approximately 15.1 grams.
Explanation:
The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want.
change 25 grams of potassium chlorate to moles.
calculate the gram molecular mass of potassium chlorate.
Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.
1 K x 39 grams/mole
+1 Cl x 35.4 grams/ mole
+3 O x 16 grams/ mole
= 122.4 grams / mole Potassium Chlorate
25
122.4
= moles.
2.05 moles of Potassium Chlorate.
There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride.
2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride.
Find the gram molecular mass of Potassium Chloride.
1 K x 39 = 39
+1 Cl x 35.4 = 35.4
= 74.4 grams / mole.
2.05 moles x 74.4 grams/ mole = 15.2 grams
Hope it helps :)
The three isomers of pentane have different structural formulas.
None. Both chlorines and both hydrogens are single-bonded to the central carbon atom; the molecule is comprised of four single bonds and no double bonds.
Hope this helps!
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
